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Read this. If the sequence is short exact, then it already has arrows $0 \to \dots$ and $\dots \to 0$ which imply injectivity and surjectivity respectively, right? So in that case there is already always such a map $t$. So what am I getting wrong?

Daniel Donnelly
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Write $$ 0 \to A \xrightarrow{q} B \xrightarrow{r} C \to 0. $$

The first and last arrows imply injectivity and surjectivity of $q$ and $r$, respectively. This does not imply that there is a splitting.

Mathworld gives a counterexample: $$ 0 \to \Bbb{Z}_2 \to \Bbb{Z}_4 \to \Bbb{Z}_2 \to 0 $$ where the first map takes $1\mapsto 2$. This is not a split SES because $\Bbb{Z}_4$ is not isomorphic to $\Bbb{Z}_2\oplus\Bbb{Z}_2$.

My geometric intuition is better than my algebraic intuition, so I tend to think of short exact sequences as analogues of fiber bundles. A split SES is special: It is like a trivializable fiber bundle. In general, one should not expect a SES to split; $B$ is generically formed by "twisting" $A$ and $C$ together.

Neal
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  • That doesn't address my question directly. – Daniel Donnelly Jun 09 '15 at 03:03
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    First paragraph addresses your assertion directly. Next two paragraphs elaborate on why your belief is wrong. If you edit an argument for your belief that injectivity/surjectivity imply splitting into the question, I will be happy to alter my answer accordingly. – Neal Jun 09 '15 at 03:30